JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 21)
A physical quantity P is described by the relation
P = a$$^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}$$ b2 c3 d$$-$$4
If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :
P = a$$^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}$$ b2 c3 d$$-$$4
If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :
8%
12%
32%
25%
Explanation
Given,
P = a$$^{{1 \over 2}}$$ b2 c3 d$$-$$4
Relative error =
$${{\Delta P} \over P}$$ $$ \times $$ 100 = ($${1 \over 2}$$ $$ \times $$ $${{\Delta a} \over a}$$ + 2$${{\Delta b} \over b}$$ + 3$${{\Delta c} \over c}$$ + 4$${{\Delta d} \over d}$$) $$ \times $$ 100
= $${1 \over 2}$$ $$ \times $$ 2 + 2 $$ \times $$ 1 + 3 $$ \times $$ 3 + 4 $$ \times $$ 5
= 32%
P = a$$^{{1 \over 2}}$$ b2 c3 d$$-$$4
Relative error =
$${{\Delta P} \over P}$$ $$ \times $$ 100 = ($${1 \over 2}$$ $$ \times $$ $${{\Delta a} \over a}$$ + 2$${{\Delta b} \over b}$$ + 3$${{\Delta c} \over c}$$ + 4$${{\Delta d} \over d}$$) $$ \times $$ 100
= $${1 \over 2}$$ $$ \times $$ 2 + 2 $$ \times $$ 1 + 3 $$ \times $$ 3 + 4 $$ \times $$ 5
= 32%
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